Let $z\in \mathbb{R}^d$.
It is claimed, easy to see, that $\displaystyle a_{\epsilon}(z)=f_{\epsilon}(|z|)\cdot \frac{z}{|z|}$ is Lipschitz continuous,i.e. there exists $C_{\epsilon}$ $$ |a_{\epsilon}(z)-a_{\epsilon}(z')|\leq |z-z'| $$ where $f_{\epsilon}(r)$ is continuous from $\mathbb{R}_+\to[0,1]$ with value $0$ when $r\leq \dfrac{\epsilon}{2}$ and value $1$ when $r>\epsilon$, in a paper.
However, I cannot see the reason. I know if both $f_{\epsilon}$ and $\dfrac{z}{|z|}$ is Lipshitz, then the proof is complete since they are both bounded. I have been attempting to find the Lipshitz coefficient for $\dfrac{z}{|z|}$ to conclude, but failed.
EDIT: In the paper, $f_{\epsilon}$ is assumed to be smooth instead of continuous. Therefore it should be Lipschitz by the mean value property.
The function $g : z \mapsto \frac{z}{|z|}$ is not Lipschitz on the whole of $\mathbb{R}^d$: take $z$ a vector of norm $1$ and consider the sequence $z_n := \left(\frac{z}{n}\right)_n$.
We then have, for $n \geq 2$: $\left|\frac{z_n}{|z_n|} - \frac{-z_n}{|- z_n|}\right| = \left|\frac{2z_n}{|z_n|}\right| = 2$ is constant yet $\left|z_n - (-z_n)\right| = 2|z_n| = \frac{2|z|}{n} = \frac{2}{n}$ tends to $0$ when $n$ tends to $\infty$.
The function $f_\varepsilon(|\cdot|)$ is what makes the function $a_\varepsilon$ Lipschitz, by "erasing" what happens towards the zero vector. $z \mapsto \frac{z}{|z|}$ is $\frac{4}{\varepsilon}$-Lipschitz (not $\frac{2}{\varepsilon}$ whoops) on each set $E_\varepsilon := \left\{z \in \mathbb{R}^d \mid |z| > \frac{\varepsilon}{2}\right\}$, and thus $a_\varepsilon$ is also Lipschitz there, at least if $f_\varepsilon$ is Lipschitz.
If $f_\varepsilon$ is only continuous I have my doubts but maybe someone else can make these doubts disappear. And then $f_\varepsilon$ being equal to $0$ on $\left\{z \in \mathbb{R}^d \mid |z| \leq \frac{\varepsilon}{2}\right\}$, $a_\varepsilon$ is also equal to $0$ there, and so should be Lipschitz everywhere.
Explication for $g : z \mapsto \frac{z}{|z|}$ being Lipschitz on the sets $E_\varepsilon$:
(Note: don't mind my change of notation for the norm, I forgot midway through. Note that my below reasoning applies to all norms, not just the euclidean one though.)
The principle of the matter is simple. For all $(x,y) \in (\mathbb{R}^d)^2$ and $(a,b) \in \mathbb{R}^2$ we have: $$\|ax - by\| \leq \|ax - ay\| + \|ay - by\| = |a| \cdot \|x - y\| + |b - a| \cdot \|y\| \tag{$*$}$$ This means that if you can bound $|a|$ by some constant $L$, and find a constant $L'$ such that $|b - a|\cdot \|y\| \leq L'\|x - y\|$, then: $$\|ax - by\| \leq (L + L')\|x - y\|$$ though that is not exactly how I'll use $(*)$ later.
Let's apply our knowledge to $a := \frac{1}{\|x\|}$ and $b := \frac{1}{\|y\|}$ for $x \in E_\varepsilon$ and any non-zero $y$: $$\begin{split} \| g(x) - g(y) \| = \|ax - by\| &\leq |a| \cdot \|x - y\| + |b - a| \cdot \|y\|\\ &\leq \frac{1}{\|x\|} \cdot \|x - y\| + \left|\frac{1}{\|x\|} - \frac{1}{\|y\|}\right| \cdot \|y\|\\ &\leq \frac{2}{\varepsilon}\|x - y\| + \left|\frac{\|y\| - \|x\|}{\|x\|\|y\|}\right| \cdot \|y\|\\ &\leq \frac{2}{\varepsilon}\|x - y\| + \left|\frac{\|y\| - \|x\|}{\|x\|}\right|\\ &\leq \frac{2}{\varepsilon}\|x - y\| + \frac{2}{\varepsilon}\Big|\|y\| - \|x\|\Big|\\ &\leq \frac{4}{\varepsilon}\|x - y\| \end{split}$$ In particular, $g$ is indeed $\frac{4}{\varepsilon}$-Lipschitz on $E_\varepsilon$.
This is also why I have some doubts as to what happens if $f_\varepsilon$ is not assumed Lipschitz:
Does there exist then a constant $L'$ such that $\left\|f_\varepsilon(\|x\|) \frac{1}{\|x\|} - f_\varepsilon(\|y\|) \frac{1}{\|y\|}\right\| \cdot \|y\| \leq L'\|x - y\|$? Yet, it's only a sufficient condition for what is desired, so even if such $L'$ doesn't exist it would not refute the original statement.
If $f_\varepsilon$ is $K$-Lipschitz, then by using $(*)$ for $x := \frac{z}{\|z\|}$, $y := \frac{z'}{\|z'\|}$, $a := f_\varepsilon(\|z\|)$ and $b := f_\varepsilon(\|z'\|)$, we obtain for $z \in E_\varepsilon$ and nonzero $z'$ (explained below): $$\left\|f_\varepsilon(\|z\|) \frac{z}{\|z\|} - f_\varepsilon(\|z'\|) \frac{z'}{\|z'\|}\right\| \leq \left(\frac{8}{\varepsilon^2} + K \right)\|z - z'\|$$ That is because: $$|a| \cdot \|x - y\| = \frac{1}{\|z\|} \cdot \|g(z) - g(z')\| \leq \frac{2}{\varepsilon} \cdot \frac{4}{\varepsilon} \|z - z'\|= \frac{8}{\varepsilon^2} \|z - z'\|$$ and: $$|b - a| \cdot \|y\| = |f_\varepsilon(\|z\|) - f_\varepsilon(\|z'\|)| \cdot 1 \leq K \Big|\|z\| - \|z'\|\Big| \leq K \|z - z'\|$$
The other reason why taking any $z'$ above is because the only remaining case to write out then is $z$ and $z'$ of norms lesser than $\frac{\varepsilon}{2}$, but $f_\varepsilon$ is zero there. So instead of separating the study in three cases (both norms greater, one lesser, both lesser) you write two of them while changing nothing. :)
And so $a_\varepsilon$ is $\frac{8}{\varepsilon^2} + K$-Lipschitz if $f$ is $K$-Lipschitz. There was probably a smarter way to go about this (which could clear my previous doubts) but at least it's better than nothing!